Find three numhers in G.P. whose sum is 35 and sum of their squares is 525.

Answer

To find three numbers in a geometric progression (G.P.) whose sum is 35 and the sum of their squares is 525, we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the three terms in G.P. be:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

### Step 2: Set up the equations based on the problem statement
From the problem, we have two conditions:
1. The sum of the terms:
\[ a + ar + ar^2 = 35 \]
This can be factored as:
\[ a(1 + r + r^2) = 35 \quad \text{(Equation 1)} \]

2. The sum of the squares of the terms:
\[ a^2 + (ar)^2 + (ar^2)^2 = 525 \]
This can be rewritten as:
\[ a^2(1 + r^2 + r^4) = 525 \quad \text{(Equation 2)} \]

### Step 3: Divide the two equations
We can divide Equation 2 by Equation 1 to eliminate \( a \):
\[ \frac{a^2(1 + r^2 + r^4)}{a(1 + r + r^2)} = \frac{525}{35} \]
This simplifies to:
\[ \frac{a(1 + r^2 + r^4)}{1 + r + r^2} = 15 \]

### Step 4: Use the factorization of \( 1 + r^2 + r^4 \)
The expression \( 1 + r^2 + r^4 \) can be factored as:
\[ (1 + r + r^2)(1 - r + r^2) \]
Thus, we can rewrite our equation as:
\[ a(1 - r + r^2) = 15 \]
Let’s call this Equation 3.

### Step 5: Solve the equations
Now we have two equations:
1. \( a(1 + r + r^2) = 35 \) (Equation 1)
2. \( a(1 - r + r^2) = 15 \) (Equation 3)

### Step 6: Isolate \( a \)
From Equation 1, we can express \( a \):
\[ a = \frac{35}{1 + r + r^2} \]
Substituting this into Equation 3 gives:
\[ \frac{35(1 - r + r^2)}{1 + r + r^2} = 15 \]
Cross-multiplying leads to:
\[ 35(1 - r + r^2) = 15(1 + r + r^2) \]
Expanding both sides:
\[ 35 - 35r + 35r^2 = 15 + 15r + 15r^2 \]
Rearranging gives:
\[ 20r^2 - 50r + 20 = 0 \]
Dividing through by 5:
\[ 4r^2 - 10r + 4 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \]
\[ r = \frac{10 \pm \sqrt{100 - 64}}{8} \]
\[ r = \frac{10 \pm \sqrt{36}}{8} \]
\[ r = \frac{10 \pm 6}{8} \]
Thus, we have:
1. \( r = 2 \)
2. \( r = \frac{1}{2} \)

### Step 8: Find \( a \)
For \( r = 2 \):
\[ a(1 + 2 + 4) = 35 \Rightarrow a \cdot 7 = 35 \Rightarrow a = 5 \]
The terms are \( 5, 10, 20 \).

For \( r = \frac{1}{2} \):
\[ a(1 + \frac{1}{2} + \frac{1}{4}) = 35 \Rightarrow a \cdot \frac{7}{4} = 35 \Rightarrow a = 20 \]
The terms are \( 20, 10, 5 \).

### Conclusion
The three numbers in G.P. are \( 5, 10, 20 \) or \( 20, 10, 5 \).